SlowGuess/NeurIPS_2023_OCR huggingface dataset:
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A Missing Proofs of Section 5
Lemma 2  $(\star)$ . Let  $(G, c)$  be a delegation graph and let  $(\mathcal{F}, E_y, y)$  be the output of Algorithm 1. Then:
(i) For every  $(G, c)$ , the output of the algorithm is unique, i.e., it does not depend on the choice of the strongly connected component in line 3.
(ii)  $\mathcal{F}$  is laminar, i.e., for any  $X,Y\in \mathcal{F}$  it holds that either  $X\subseteq Y$ $Y\subseteq X$  , or  $X\cap Y = \emptyset$
(iii) Branching  $B$  in  $(G,c)$  is min-cost iff (a)  $B\subseteq E_y$  , and (b)  $|B\cap \delta^{+}(X)| = 1$  for all  $X\in \mathcal{F},X\subseteq N$
(iv) For every  $X \in \mathcal{F}$ , an in-tree  $T$  in  $G[X] = (X, E[X])$ , where  $E[X] = \{(u, v) \in E \mid u, v \in X\}$ , is min-cost iff  $(a) T \subseteq E_y$ , and  $(b) |T \cap \delta^+(Y)| = 1$  for all  $Y \in \mathcal{F}$  such that  $Y \subset X$ .
Proof. Let  $(G, c)$  be a delegation graph and let  $(\mathcal{F}, E_y, y)$  be the output of Algorithm 1.
We start by proving statement (ii). The sets in  $\mathcal{F}$  correspond exactly to those sets with positive  $y$ -value. Assume for contradiction that there exist two sets  $X, Y \in \mathcal{F}$  with  $X \cap Y \neq \emptyset$  and none of the subsets is a subset of the other. Assume without loss of generality that  $X$  was selected before  $Y$  by the algorithm and let  $y_1$  and  $y_2$  be the status of the function  $y$  in each of the two situations. Then, by construction of the algorithm it holds that  $G_1 = (N \cup S, E_{y_1})$  is a subgraph of  $G_2 = (N \cup S, E_{y_2})$ . This is because once an edge is added to the set of tight edges (denoted by  $E_y$ ) it remains in this set. Since  $Y$  is a strongly connected component in  $G_2$  without outgoing edge, it holds that for every  $z \in X \setminus Y$  and  $z' \in X \cap Y$ , the node  $z'$  does not reach  $z$  in  $G_2$ . However, this is a contradiction to the fact that  $X$  is a strongly connected component in the graph  $G_1$ , which concludes the proof of statement (ii).
We now turn to prove statement (i) and already assume that  $\mathcal{F}$  is laminar. We fix an order of the selected strongly connected components in line 3 of the algorithm. Then, suppose that for some other choices in line 3, the algorithm returns some other output  $(\hat{\mathcal{F}}, E_{\hat{y}}, \hat{y})$ . Note that  $\hat{\mathcal{F}} \neq \mathcal{F}$  or  $E_{\hat{y}} \neq E_y$  implies that  $\hat{y} \neq y$ . Thus, it suffices to assume for contradiction that  $\hat{y} \neq y$ . Then there must be a smallest set  $X$ , that has  $y(X) \neq \hat{y}(X)$  (without loss of generality we assume  $y(X) > \hat{y}(X)$ ). Let  $\mathcal{X} = 2^X \setminus \{X\}$  be the set of all strict subsets of  $X$ . Since we defined  $X$  to be of minimal cardinality, we have  $y[\mathcal{X}] = \hat{y}[\mathcal{X}]$ , where  $y[\mathcal{X}]$ , and  $\hat{y}[\mathcal{X}]$  denote the restriction of  $y$  and  $\hat{y}$  to  $\mathcal{X}$ , respectively. Because  $y(X) > 0$ , all children of  $X$  are strongly connected by tight edges with respect to  $y[\mathcal{X}]$  and have no tight edges pointing outside of  $X$ . Now, consider the iteration of the alternative run of the Algorithm 2, in which the algorithm added the last set in  $\mathcal{X} \cup \{X\}$ . Since  $\hat{y}(X) < y(X)$ , for every further iteration of the algorithm, a chosen set  $X' \neq X$  cannot contain any node in  $X$  (because otherwise  $X'$  cannot form a strongly connected component without outgoing edge). However, since the nodes in  $X$  cannot reach a sink via tight edges, this is a contradiction to the termination of the algorithm.
We now prove statement (iii). The plan of attack is the following: First we define a linear program that captures the min-cost branchings in a delegation graph. Second, we dualize the linear program and show that  $y$  (more precisely a minor variant of  $y$ ) is an optimal solution to the dual LP, and third, utilize complementary slackness to prove the claim. For a given delegation graph  $(G,c)$  with  $V(G) = N \cup S$  we define the following linear program, also denoted by (LP):
$$
\begin{array}{l} \min  \sum_ {e \in E} c (e) x _ {e} \\ \sum_ {e \in \delta^ {+} (X)} x _ {e} \geq 1 \quad \forall X \subseteq N \\ x _ {e} \geq 0 \quad \forall e \in E \\ \end{array}
$$
We claim that every branching  $B$  in  $G$  induces a feasible solution to (LP). More precisely, given a branching  $B$ , let
$$
x _ {e} = \left\{ \begin{array}{l l} 1 & \text {i f} e \in B \\ 0 & \text {i f} e \notin B. \end{array} \right.
$$
The last constraint is trivially satisfied. Now, assume for contradiction that there exists  $X \subseteq N$  such that the corresponding constraint in (LP) is violated. In this case the nodes in  $X$  have no path towards some sink node in  $B$ , a contradiction to the fact that  $B$  is a (maximum cardinality) branching. In particular, this implies that the objective value of (LP) is at most the minimum cost of any branching in  $G$  (in fact the two values are equal, but we do not need to prove this at this point). We continue by deriving the dual of (LP), to which we refer to as (DLP):
$$
\max  \sum_ {X \subseteq N} y _ {X}
$$
$$
\sum_ {X \subseteq N | e \in \delta^ {+} (X)} y _ {X} \leq c (e) \quad \forall e \in E
$$
$$
y _ {X} \geq 0 \quad \forall X \subseteq N
$$
Now, let  $y$  be the function returned by Algorithm 1. We define  $\hat{y}$ , which is intuitively  $y$  restricted to all subsets on  $N$ , more precisely,  $\hat{y}(X) = y(X)$  for all  $X \subseteq N$ . We claim that  $\hat{y}$  is a feasible solution to (DLP). This can be easily shown by induction. More precisely, we fix any  $e \in E$  and show that the corresponding constraint in (DLP) is satisfied throughout the execution of the algorithm. At the beginning of the algorithm  $y$  (and hence  $\hat{y}$ ) is clearly feasible for (DLP). Now, consider any step in the algorithm and let  $X$  be the selected strongly connected component. If  $e \in \delta^{+}(X)$ , then we know that the constraint corresponding to  $e$  is not tight (since  $X$  has no tight edge in its outgoing cut). Moreover,  $y$  is only increased up to the point that some edge in  $\delta^{+}(X)$  becomes tight (and not higher than that). Hence, after this round, the constraint for  $e$  is still satisfied. If, on the other hand,  $e \notin \delta^{+}(X)$ , then the left-hand-side of  $e$ 's constraint remains equal when  $y(X)$  is increased. Hence, the constraint of  $e$  is still satisfied.
Next, we claim that there exists a branching  $B$  in  $G$ , such that for the resulting primal solution  $x$ , it holds that  $\sum_{e \in E} c(e)x_e = \sum_{X \subseteq N} \hat{y}(X)$ . The branching  $B$  will be constructed in a top-down fashion by moving along the laminar hierarchy of  $\mathcal{F}$ . To this end let  $G_X$  be the contracted graph as defined in Algorithm 2. We start by setting  $X = N \cup S$ . Since every node in  $N$  can reach some sink via tight edges, we also know that every node in  $G_X$  can reach some sink. Hence, a branching in  $G_X$  has exactly one edge per node in  $V_X$  that is not a sink. Let's pick such a branching  $B_X$ . We know that for every edge in  $B_X = (Y,Z)$  there exists some edge in the original graph  $G$  that is also tight, i.e.,  $u \in Y$  and  $v \in Z$  such that  $(u,v) \in E_y$ . For every edge in  $B_X$  pick an arbitrary such edge and add it to  $B$ . Now, pick an arbitrary node  $Y \in V(G_X)$ . By construction, we know that exactly one edge from  $B$  is included in  $\delta^+(Y)$ , call this edge  $(u,v)$ . Then, within the graph  $G_Y$ , there exists exactly one node  $Z \in V(G_Y)$ , that contains  $u$ . We are going to search for a  $Z$ -tree within  $G_Y$ . We know that such a tree exists since  $G_Y$  is strongly connected by construction. We follow the pattern from before, i.e., finding a  $Z$ -tree, mapping the edges back to the original graph (arbitrarily), and then continuing recursively. For proving our claim, it remains to show that  $\sum_{e \in E} c(e)x_e = \sum_{X \subseteq N} \hat{y}(X)$ . The crucial observation is that, by construction, every set in  $\hat{\mathcal{F}} = \mathcal{F} \setminus \{s | s \in S\}$  is left by exactly one edge in  $B$ . Hence, we can partition the set  $\hat{\mathcal{F}}$  into sets  $\bigcup_{e \in B} \hat{F}_e$ , where  $\hat{\mathcal{F}}_e = \{X \in \hat{\mathcal{F}} | e \in \delta^+(X)\}$ . Moreover, observe that every edge in  $B$  is tight. As a result we get that
$$
\sum_ {e \in B} c (e) x _ {e} = \sum_ {e \in B} \sum_ {X \in \hat {\mathcal {F}} _ {e}} \hat {y} (X) = \sum_ {X \subseteq N} \hat {y} (X),
$$
proving the claim.
As a result, note that we found a primal solution  $B$  (precisely, the  $x$  induced by  $B$ ), and a dual solution  $\hat{y}$  having the same objective value. By weak duality, we can conclude that both solutions are in particular optimal. It only remains to apply complementary slackness to conclude the claim. To this end, let  $B$  be a min-cost branching and  $x$  be the induced primal solution. By the argument above we know that  $x$  is optimal. Now, for any  $X \subseteq N$  for which  $\hat{y}(X) > 0$  (hence  $X \in \mathcal{F}$ ), complementary slackness prescribes that the corresponding primal constraint is tight, i.e.,  $\sum_{e \in \delta^{+}(X)} x_{e} = 1$ . Hence, the branching corresponding to  $x$  leaves the set  $X$  exactly once, and part (b) of statement (iii) is
satisfied. For statement (a) we apply complementary slackness in the other direction. That is, when  $x_{e} > 0$ , this implies that the corresponding dual constraint is tight, implying that  $e$  has to be tight with respect to  $\hat{y}$  and therefore also with respect to  $y$  (recall that  $y$  and  $\hat{y}$  only differ with respect to the sink nodes).
We now turn to proving statement (iv). This is done almost analogously to statement (iii). Fix  $X \in \mathcal{F}$ . In the following we argue about the min-cost in-trees in  $G[X]$  and how to characterize these via a linear program. To this end, we add a dummy sink node  $r$  to the graph  $G[X]$  and call the resulting graph  $\hat{G}$ . More precisely,  $\hat{G} = (X \cup \{r\}, E[X] \cup \{(u, r) \mid u \in X\})$ . The cost of any edge  $(u, r), u \in X$  is set to  $c^* := \max_{e \in E(G)} c(e) + 1$ , where it is only important that this value is larger than any other cost in the graph. We define the following LP:
$$
\min  \sum_ {e \in E (\hat {G})} c (e) x _ {e}
$$
$$
\sum_ {e \in \delta_ {\mathcal {G}} ^ {+} (Z)} x _ {e} \geq 1 \quad \forall Z \subseteq X
$$
$$
x _ {e} \geq 0 \quad \forall e \in E (\hat {G})
$$
For every min-cost in-tree  $T$  in  $G[X]$  we obtain a feasible solution to (LP). To this end, let  $u \in X$  be the sink node of  $T$  and define  $\hat{T} = T \cup \{(u,r)\}$ . Then, translate  $\hat{T}$  to its incidence vector  $x$ . Given this observation, we again derive the dual of (LP), to which we refer to as (DLP):
$$
\max  \sum_ {Z \subseteq X} y _ {Z}
$$
$$
\sum_ {Z \subseteq X | e \in \delta_ {\hat {G}} ^ {+} (Z)} y _ {Z} \leq c (e) \quad \forall e \in E (\hat {G})
$$
$$
y _ {Z} \geq 0 \quad \forall Z \subseteq X
$$
Now, let  $y$  be the output of Algorithm 1 for the original graph  $G$ . We derive  $\hat{y}:2^{X}\to \mathbb{R}$  as follows:
$$
\hat {y} (Z) = \left\{ \begin{array}{l l} y (Z) & \text {i f} Z \subset X \\ c ^ {*} - \max  _ {u \in X} \sum_ {Z \subset X | e \in \delta_ {\mathcal {G}} ^ {+} (Z)} y (Z) & \text {i f} Z = X \end{array} \right.
$$
First, analogously to (iii), it can be verified that  $\hat{y}$  is a feasible solution to (DLP). Moreover, again analogously to (iii), there exists some min-cost  $r$ -tree in  $\hat{G}$  and a corresponding primal solution  $x$ , such that  $\sum_{e \in E(\hat{G})} c(e)x_e = \sum_{Z \subseteq X} \hat{y}_X$ . (This tree is derived by first choosing a tight edge towards the dummy root node  $r$  and then again recurse over the laminar family  $\mathcal{F}$  restricted to  $X$ .) This implies by weak duality that  $\hat{y}$  is an optimal solution to (DLP) and any min-cost  $r$ -tree in  $\hat{G}$  is an optimal solution to (LP). As a result, we can again apply complementary slackness in both directions: Let  $T$  be a min-cost in-tree in  $G[X]$  with sink node  $u \in X$ . Then let  $\hat{T} = T \cup \{(u, r)\}$  be the corresponding min-cost  $r$ -tree in  $\hat{G}$  and  $x$  be the corresponding incidence vector. Then, complementary slackness implies that for any  $e \in E[X]$  for which  $x_e > 0$  (and hence  $e \in T$ ), it holds that the corresponding constraint in (DLP) is tight with respect to  $\hat{y}$  (and also  $y$ ). This implies that  $e \in E_y$ . On the other hand, for any  $Z \subset X$ , if  $\hat{y}_Z > 0$ , and hence  $X \in \mathcal{F}$ , complementary slackness prescribes that the corresponding primal constraint is tight, and hence  $|T \cap \delta_{G[X]}^+(Z)| = 1$ , concluding the proof.
For the proof of the next theorem, we first explain how to compute the absorbing probabilities of an absorbing Markov chain  $(G,P)$  and show a related lemma that we need in Appendix C. W.l.o.g. we assume that the states  $V(G)$  are ordered such that the non-absorbing states  $N$  come first and the absorbing states  $S$  last. We can then write the transition matrix as
$$
P = \left[ \begin{array}{c c} D & C \\ 0 & I _ {| S |} \end{array} \right] \quad ,
$$
where  $D$  is the  $|N|\times |N|$  transition matrix from non-absorbing states to non-absorbing states and  $C$  is the  $|N|\times |S|$  transition matrix from non-absorbing states to absorbing states.  $I_{|S|}$  denotes the  $|S|\times |S|$  identity matrix. The absorbing probability of an absorbing state  $s\in S$ , when starting a random walk in a state  $v\in N$  is then given as the entry in the row corresponding to  $v$  and the column corresponding to  $s$  in the  $|N|\times |S|$  matrix  $(I_{|N|} - D)^{-1}C$  [Grinstead and Snell, 1997].
Lemma A.1. Adding a self-loop to a non-absorbing state  $v$  with probability  $p$  and scaling all other transition probabilities from that state by  $1 - p$  does not change the absorbing probabilities of an absorbing Markov-chain  $(G, P)$ .
Proof. Let  $(D, C)$  and  $(D', C')$  be the transition matrices of the absorbing Markov chain before and after adding the self-loop. Let  $\mathbf{d}_v, \mathbf{d}_v', \mathbf{c}_v, \mathbf{c}_v'$  be the rows of  $D, D', C, C'$ , corresponding to state  $v$  respectively. Then
$$
\mathbf {d} _ {v} ^ {\prime} = (1 - p) \mathbf {d} _ {v} + p \mathbf {e} _ {v} ^ {\intercal},
$$
$$
\mathbf {c} _ {v} ^ {\prime} = (1 - p) \mathbf {c} _ {v}
$$
and  $\mathbf{d}_u^{\prime} = \mathbf{d}_u$  and  $\mathbf{c}_u^\prime = \mathbf{c}_u$  for all  $u\neq v$
We want to show that  $(I_{|N|} - D)^{-1}C = (I_{|N|} - D')^{-1}C'$ . Let  $Z = (I_{|N|} - D)^{-1}C$ . Then  $Z = (I_{|N|} - D')^{-1}C'$  if and only if  $Z = D'Z + C'$ . Notice, that only the row corresponding to  $v$  in  $D'$  and  $C'$  differ from  $D$  and  $C$  and therefore for all  $u \neq v$
$$
\mathbf {z} _ {u} = \mathbf {d} _ {u} Z + \mathbf {c} _ {u} = \mathbf {d} _ {u} ^ {\prime} Z + \mathbf {c} _ {u} ^ {\prime},
$$
where  $\mathbf{z}_u$  is the row of  $Z$  corresponding to  $u$ . The only thing left to show is  $\mathbf{z}_v = \mathbf{d}_v' Z + \mathbf{c}_v'$ . We have
$$
\begin{array}{l} \mathbf {d} _ {v} ^ {\prime} Z + \mathbf {c} _ {v} ^ {\prime} = ((1 - p) \mathbf {d} _ {v} + p \mathbf {e} _ {v} ^ {\intercal}) Z + (1 - p) \mathbf {c} _ {v} \\ = (1 - p) \mathbf {d} _ {v} Z + p \mathbf {e} _ {v} ^ {\intercal} Z + (1 - p) \mathbf {c} _ {v} \\ = (1 - p) \left(\mathbf {d} _ {v} Z + \mathbf {c} _ {v}\right) + p \mathbf {e} _ {v} ^ {\intercal} Z \\ = (1 - p) \mathbf {z} _ {v} + p \mathbf {z} _ {v} \quad (\text {s i n c e} D Z + C = Z) \\ = \mathbf {z} _ {v}, \\ \end{array}
$$
Theorem 4  $(\star)$ . Algorithm 2 returns MIXED BORDA BRANCHING and runs in poly(n).
Proof. We start by showing by induction that the given interpretation of the weight function on the nodes is correct, i.e., for any  $v \in N$ ,  $t_X(v)$  corresponds to the number of min-cost  $v$ -trees in the graph  $G[X]$ . The claim is clearly true for any singleton, since  $t_{\{v\}}(v) = 1$  and the number of  $v$ -trees in  $(\{v\}, \emptyset)$  is one, i.e., the empty set is the only  $v$ -trees. Now, we fix some  $X \in \mathcal{F}'$  and assume that the claim is true for all children of  $X$ . In the following, we fix  $v \in X$  and argue that the induction hypothesis implies that the claim holds for  $t_X(v)$  as well.
For any node  $u \in X$ , let  $Y_u \in \mathcal{F}$  be the child of  $X$  containing node  $u$ . Moreover, let  $\mathcal{T}_v^*(G[X])$  (or short  $\mathcal{T}_v^*)$  be the set of min-cost  $v$ -trees in  $G[X]$ , and  $\mathcal{T}_{Y_v}(G_X)$  (or short  $\mathcal{T}_{Y_v}$ ) be the set of  $Y_v$ -trees in  $G_X$ . Lastly, for any  $u \in X$ , let  $\mathcal{T}_u^*(G[Y_u])$  be the set of min-cost  $u$ -trees in  $G[Y_u]$ . In the following, we argue that there exists a many-to-one mapping from  $\mathcal{T}_v^*$  to  $\mathcal{T}_{Y_v}$ . Note that, by statement (iv) in Lemma 2, every min-cost in-tree  $T$  in  $G[X]$  (hence,  $T \in \mathcal{T}_v^*)$  leaves every child of  $X$  exactly once via a tight edge. Therefore, there exists a natural mapping to an element of  $\mathcal{T}_{Y_v}$  by mapping every edge in  $T$  that connects two children of  $X$  to their corresponding edge in  $G_X$ . More precisely,  $\hat{T} = \{(Y, Y') \in E_X \mid T \cap \delta^+(Y) \cap \delta^-(Y') \neq \emptyset\}$  is an  $Y_v$ -tree in  $G_X$  and hence an element of  $\mathcal{T}_{Y_v}$ .
Next, we want to understand how many elements of  $\mathcal{T}_v^*$  map to the same element in  $\mathcal{T}_{Y_v}$ . Fix any  $\hat{T} \in \mathcal{T}_{Y_v}$ . We can construct elements of  $\mathcal{T}_v^*$  by combining (an extended version of)  $\hat{T}$  with min-cost in-trees within the children of  $X$ , i.e., with elements of the sets  $\mathcal{T}_u^*(G[Y_u])$ ,  $u \in X$ . More precisely, for any edge  $(Y, Y') \in \hat{T}$ , we can independently choose any of the edges in  $(u, u') \in E_y \cap (Y \times Y')$  and combine it with any min-cost  $u$ -tree in the graph  $G[Y]$ . This leads to
$$
\Big(\prod_{(Y,Y^{\prime})\in \hat{T}}\sum_{(u,u^{\prime})\in E_{y}\cap (Y\times Y^{\prime})}t_{Y}(u)\Big)t_{Y_{v}}(v) = \Big(\prod_{(Y,Y^{\prime})\in \hat{T}}w_{X}(Y,Y^{\prime})\Big)\cdot t_{Y_{v}}(v)
$$
many different elements from  $\mathcal{T}_v^*$  that map to  $\hat{T}\in \mathcal{T}_{Y_v}$ . Hence,
$$
| \mathcal {T} _ {v} ^ {*} | = \sum_ {\hat {T} \in \mathcal {T} _ {Y _ {v}}} \prod_ {(Y, Y ^ {\prime}) \in \hat {T}} w _ {Y} (Y, Y ^ {\prime}) \cdot t _ {Y _ {v}} (v) = w _ {X} (\mathcal {T} _ {Y _ {v}}) \cdot t _ {Y _ {v}} (v) = t _ {X} (v),
$$
where the last inequality follows from the definition of  $t_X(v)$  in the algorithm. This proves the induction step, i.e.,  $t_X(v)$  corresponds to the number of min-cost  $v$ -trees in the graph  $G[X]$ .